我们给了两个链表，任务是使用链表的整数元素形成一对，以使它们的乘积等于给定值，即k。链表是一系列数据结构，它们通过链接连接在一起。
输入项
vector v_1 = {5, 7, 8, 10, 11},.
vector v_2 = {6, 4, 3, 2, 0} , int k = 20输出结果
Count of pairs from two linked lists whose product is equal to a given value k are: 2
说明
The pairs which can be formed using the given linked lists are: (5, 6) = 30(not equals to k), (5, 4) = 20(equals to k), (5, 3) = 15(not equals to k), (5, 2) = 10(not equals to k), (5, 0) = 0(not equals to k), (7, 6) = 42(not equals to k), (7, 4) = 28(not equals to k), (7, 3) = 21(not equals to k), (7, 2) = 14(not equals to k), (7, 0) = 0(not equals to k), (8, 6) = 48(not equals to k), (8, 4) = 32(not equals to k), (8, 3) = 24(not equals to k), (8, 2) = 16(not equals to k), (8, 0) = 0(not equals to k), (10, 6) = 60(not equals to k), (10, 4) = 40(not equals to k), (10, 3) = 30(not equals to k), (10, 2) = 20(not equals to k), (10, 0) = 0(not equals to k), (11, 6) = 66(not equals to k), (11, 4) = 44(not equals to k), (11, 3) = 3(not equals to k), (11, 2) = 22(not equals to k), (11, 0) = 0(not equals to k). So, clearly there are 2 pairs which are equal to the given product.
输入项
vector v_1 = {2, 3, 5, 6},.
vector v_2 = {6, 4, 3} , int k = 9输出结果
Count of pairs from two linked lists whose sum is equal to a given value k are: 1
说明
The pairs which can be formed using the given linked lists are: (2, 6) = 12(not equals to k), (2, 4) = 8(not equals to k), (2, 3) = 6(not equals to k), (3, 6) = 18(not equals to k), (3, 4) = 12(not equals to k), (3, 3) = 9(equals to k), (5, 6) = 30(not equals to k), (5, 4) = 20(not equals to k), (5, 3) = 15(not equals to k), (6, 6) = 36(not equals to k), (6, 4) = 24(not equals to k), (6, 3) = 18(not equals to k),. So, clearly there is 1 pair which is equal to the given sum.
将k的值和整数类型的值输入到两个向量中，这样我们就可以传递向量以形成一个链表
创建一个函数，该函数将使用作为参数传递给函数的向量创建一个链表。
遍历循环直至向量的大小，并创建类的指针对象
列表节点
在ptr-> next不等于NULL的情况下遍历循环，并将ptr设置为ptr-> next
在ptr-> next集合矢量[i]内部
返回开始
创建一个函数，该函数将返回与给定产品匹配的对数。
进行临时变量计数并将其设置为0
创建两个指针对象，即，* first_list用于第一个链接列表，* second_list用于第二个链接列表。
从第一个列表的开始指针开始循环，直到列表不为空
在循环内部，从第二个列表的开始指针开始另一个循环，直到列表不为空
在循环内，检查IF（first_list-> data * second_list-> data）== k，然后将计数加1
返回计数
打印结果。
#include<bits/stdc++.h>
using namespace std;
class ListNode{
public:
int data;
ListNode *next;
ListNode(int data){
this->data = data;
next = NULL;
}
};
ListNode *CreateList(vector v){
ListNode *start = new ListNode(v[0]);
for (int i = 1; i < v.size(); i++){
ListNode *ptr = start;
while (ptr->next != NULL){
ptr = ptr->next;
}
ptr->next = new ListNode(v[i]);
}
return start;
}
int product_pair(ListNode *start_1, ListNode *start_2, int k){
int count = 0;
ListNode *first_list , *second_list;
for (first_list = start_1; first_list != NULL; first_list = first_list->next){
for (second_list = start_2; second_list != NULL; second_list = second_list->next){
if ((first_list->data * second_list->data) == k){
count++;
}
}
}
return count;
}
int main(){
vector<int> v_1 = {5, 7, 8, 10, 11};
ListNode* start_1 = CreateList(v_1);
vector v_2 = {6, 4, 3, 2, 0};
ListNode* start_2 = CreateList(v_2);
int k = 30;
cout<<"Count of pairs from two linked lists whose product is equal to a given value k are: "<<product_pair(start_1, start_2, k);
}输出结果
如果我们运行上面的代码，它将生成以下输出-
Count of pairs from two linked lists whose product is equal to a given value k are: 2